Answer :
To determine the volume of carbon dioxide (CO₂) gas released at standard temperature and pressure (STP) when 6.00 grams of carbon is burned completely, follow these steps:
1. Identify the balanced chemical equation:
The combustion of carbon can be represented by the equation:
[tex]\[ \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \][/tex]
This indicates that 1 mole of carbon (C) combines with 1 mole of oxygen (O₂) to produce 1 mole of carbon dioxide (CO₂).
2. Calculate the molar mass of carbon (C):
The molar mass of carbon (C) is:
[tex]\[ 12.01 \text{ grams per mole} \][/tex]
3. Determine the number of moles of carbon in 6.00 grams:
Use the formula:
[tex]\[ \text{moles of carbon} = \frac{\text{mass of carbon}}{\text{molar mass of carbon}} \][/tex]
Plugging in the values:
[tex]\[ \text{moles of carbon} = \frac{6.00 \text{ grams}}{12.01 \text{ grams per mole}} \approx 0.4996 \text{ moles} \][/tex]
4. Understand the stoichiometry of the reaction:
According to the balanced equation, 1 mole of carbon produces 1 mole of CO₂. Therefore, 0.4996 moles of carbon will produce the same number of moles of CO₂.
5. Calculate the volume of CO₂ gas produced at STP:
At standard temperature and pressure (STP), 1 mole of any ideal gas occupies 22.4 liters.
[tex]\[ \text{volume of CO}_2 = \text{moles of CO}_2 \times 22.4 \text{ liters per mole} \][/tex]
Substituting the values:
[tex]\[ \text{volume of CO}_2 = 0.4996 \times 22.4 \approx 11.19 \text{ liters} \][/tex]
6. Determine the closest volume option:
By comparing the calculated volume (11.19 liters) with the given options:
[tex]\[ 6.0 \text{ L}, \quad 11.2 \text{ L}, \quad 22.4 \text{ L}, \quad 134.4 \text{ L} \][/tex]
The closest value to 11.19 liters is 11.2 liters.
Therefore, when 6.00 grams of carbon is burned completely, the volume of carbon dioxide gas released at STP is 11.2 liters.
1. Identify the balanced chemical equation:
The combustion of carbon can be represented by the equation:
[tex]\[ \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \][/tex]
This indicates that 1 mole of carbon (C) combines with 1 mole of oxygen (O₂) to produce 1 mole of carbon dioxide (CO₂).
2. Calculate the molar mass of carbon (C):
The molar mass of carbon (C) is:
[tex]\[ 12.01 \text{ grams per mole} \][/tex]
3. Determine the number of moles of carbon in 6.00 grams:
Use the formula:
[tex]\[ \text{moles of carbon} = \frac{\text{mass of carbon}}{\text{molar mass of carbon}} \][/tex]
Plugging in the values:
[tex]\[ \text{moles of carbon} = \frac{6.00 \text{ grams}}{12.01 \text{ grams per mole}} \approx 0.4996 \text{ moles} \][/tex]
4. Understand the stoichiometry of the reaction:
According to the balanced equation, 1 mole of carbon produces 1 mole of CO₂. Therefore, 0.4996 moles of carbon will produce the same number of moles of CO₂.
5. Calculate the volume of CO₂ gas produced at STP:
At standard temperature and pressure (STP), 1 mole of any ideal gas occupies 22.4 liters.
[tex]\[ \text{volume of CO}_2 = \text{moles of CO}_2 \times 22.4 \text{ liters per mole} \][/tex]
Substituting the values:
[tex]\[ \text{volume of CO}_2 = 0.4996 \times 22.4 \approx 11.19 \text{ liters} \][/tex]
6. Determine the closest volume option:
By comparing the calculated volume (11.19 liters) with the given options:
[tex]\[ 6.0 \text{ L}, \quad 11.2 \text{ L}, \quad 22.4 \text{ L}, \quad 134.4 \text{ L} \][/tex]
The closest value to 11.19 liters is 11.2 liters.
Therefore, when 6.00 grams of carbon is burned completely, the volume of carbon dioxide gas released at STP is 11.2 liters.
